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Effectus [21]
3 years ago
5

A man with normal lungs and an arterial PO2 of 40 mmHg takes an overdose of barbiturates that halves his alveolar ventilation wi

thout changing his metabolism. If his respiratory exchange ratio is 0.8, how much does his inspired oxygen concentration (%) have to be increased to return his alveolar PO2 to the original level?
Chemistry
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

\frac{4}{5} = \frac{40}{x}

Where x represent the value of interest on this case. And solving for the value of x we have:

x = 40 * \frac{5}{4}= 50 mm Hg

So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.

And in order to find the concentration we can use a figure called the "O2

-CO2  diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%  

So then the final answer for this case would be an increase of 7%

Explanation:

For this case we know that a man with normal lungs have an arterial Po2 os 40 mm Hg.

Then we know that this man take an overdose of barbiturates thats halves his aveolar ventilation without changing his metabolism

We also know that the respiratory exchange ratio is 0.8 or 8/10

0.8 = \frac{8}{10}= \frac{4}{5}

And we want to find hor much does his inspired oxyden concentration % have to increased to return his avelolar Po2 to the original level.

On this case we can apply a proportion rule and we have this:

\frac{4}{5} = \frac{40}{x}

Where x represent the value of interest on this case. And solving for the value of x we have:

x = 40 * \frac{5}{4}= 50 mm Hg

So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.

And in order to find the concentration we can use a figure called the "O2

-CO2  diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%  

So then the final answer for this case would be an increase of 7%

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emmainna [20.7K]

Answer:

(C) is the mechanism that increases the temperature between the photosphere and corona of the sun.

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When a stars is converted from hydrogen to helium then this is known as proton -proton chain reaction.Due to this reaction the temperature between corona of the sun and photosphere increases.The main reason for increasing the temperature is exothermic reaction.

So the option C is correct.

(C) is the mechanism that increases the temperature between the photosphere and corona of the sun.

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2 propanoate on oxidation gives ?
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Answer:

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7 0
3 years ago
How many grams of C6H6O3 will react with 97.0grams of O2?<br><br> C6H6O3+3O2 ===&gt; 6CO+3H2O
likoan [24]

Answer:

127.3 g

Explanation:

Data Given

mass of O₂= 97 g

mass of C₆H₆O₃ = ?

Reaction Given:

                C₆H₆O₃ + 3O₂ -------->  6CO + 3H₂O

Solution:

First find the mass of C₆H₆O₃ from the reaction that it combine with how many grams of oxygen (O₂).

Look at the balanced reaction

                  C₆H₆O₃  +   3O₂       -------->  6CO + 3H₂O

                    1 mol        3 mol

So 1 mole of C₆H₆O₃ combine with 3 moles of O₂  

Now

convert the moles into mass for which we have to know molar mass of C₆H₆O₃ and O₂

Molar mass of C₆H₆O₃

Molar mass of C₆H₆O₃ = 6(12) + 6(1) +3(16) = 72 + 6 + 48

Molar mass of C₆H₆O₃ = 126 g/mol

mass of C₆H₆O₃

                 mass in grams = no. of moles x molar mass

                 mass of C₆H₆O₃ = 1 mol x 126 g/mol

                 mass of C₆H₆O₃ = 126 g

Molar mass of O₂ = 2(16)

Molar mass of O₂= 32 g/mol

mass of O₂

                mass in grams = no. of moles x molar mass

                mass of O₂ = 3 mol x 32 g/mol

                 mass of O₂ = 96 g

So,

96 g of O₂ combine with 126 g of C₆H₆O₃ then how many grams of C₆H₆O₃ will combine with 97.0grams of O2

Apply unity Formula

                      96 g of O₂ ≅ 126 g of C₆H₆O₃

                      97 g of O₂ ≅  X g of C₆H₆O₃

By cross multiplication

                    X g of O₂ = 126 g x 97 g / 96 g

                   X g  of O₂ = 127.3 g

127.3 g of C₆H₆O₃ will combine with 97 g of O₂

So

mass of C₆H₆O =  127.3 g

7 0
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Answer:

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Explanation:

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